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22483Re: CONJECTURE with PRIZE :-)

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  • Antreas Hatzipolakis
    Jul 1 9:33 AM


      [APH]:

      I offer an old book in German on Triliner Coordinates.

      Well .... it is available on-line, but it is an other thing to have in your
      library the original copy and other thing to have an electronic copy :-)

      Now about the conjecture:

      A way to prove it is to find the locus of points with
      the property in question.

      That is:

      Let ABC be a triangle and P a point.
      Denote:
      La, Lb, Lc = the Euler lines of PBC,PCA,PAB, resp.
      Na, Nb, Nc = the NPC centers of PBC,PCA, PAB, resp.
      The perpendicular to La at Na intersects BC,CA,AB at Aa, Ab,Ac,
      resp.
      Denote:
      A' = BAb /\ CAc. Similarly B',C'.

      Which is the locus of P such that ABC, A'B'C' are perspective?
      (or, equivalently, Aa, Bb, Cc are collinear?)

      If part of the locus is the Neuberg Cubic, then the conjecture
      is proved true !


      **************************************************

      [Peter Moses]:

      Hi Antreas,
       
      Circumcircle + Infinity + Neuberg Cubic, cyclicsum[a^2 ((S^2 - 3 SA SB) y^2 z - (S^2 - 3 SC SA) y z^2)], + maybe 3 imaginary ellipses, (-a^2 + b^2 + c^2) y z + c^2 y^2 + b^2 z^2 = 0, centered on the vertices of ABC ?
       
      Best regards,
      Peter.



      *******************************************

      Comments:

      1. Thanks, Peter !!!
      The book is yours ..... :-)

      2. A question is: which is the locus of the points of concurrence
      when P moves on the Neuberg cubic?
      How about simple simple P's: Orthocenter, Incenter, Fermat's .... ?
      Are the points of concurrence listed in ETC?

      3. Whenever I see a theorem / problem, which looks complicated,
      I am wondering how the author found (composed) it!
      The same may some of you think about this problem.
      Why the NPC centers? you may ask.
      Well.... let's take the orthocenter H and the Euler lines La,Lb,Lc
      and the NPC centers Na,Nb,Nc of HBC, HCA,HAB, resp.

      We have:
      La,Lb,Lc = AN,BN,CN, resp. and Na,Nb,Nc = N
      On the other hand we know that for any point P,
      the perpendiculars to AP,BP,CP at P intersect
      BC,CA,AB, resp. at three collinear points
      (it is an old theorem by Brocard, equivalent to perspective variation.
      There were discussions in Hyacinthos).
      Now, the generalization is obvious (since the locus of
      P such that the Euler lines of PBC,PCA,PAB concur is the Neuberg cubic
      + (O) + Linf.).

      Antreas
       
       




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